Let $f$ be a vector-valued function defined by $f(t)=(\ln(4t),3\log_2(t))$. Find $f'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{4t}+\dfrac{1}{3t\ln(2)}$ (Choice B) B $\left(\dfrac{1}{4t},-\dfrac{3}{2t}\right)$ (Choice C) C $\left(\dfrac{1}{4t},\dfrac{3}{t\ln(2)}\right)$ (Choice D) D $\left(\dfrac{1}{t},\dfrac{3}{t\ln(2)}\right)$
$f$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $f(t)=(\ln(4t),3\log_2(t))$. Let's differentiate the first expression: $\begin{aligned}\dfrac{d}{dt}[\ln(4t)]&=\dfrac{d}{dt}[\ln(4)+\ln(t)] \\\\&=\dfrac{1}{t}\end{aligned}$ Let's differentiate the second expression: $\dfrac{d}{dt}[3\log_2(t)]=\dfrac{3}{t\ln(2)}$ Now let's put everything together: $\begin{aligned} f'(t)&=\left(\dfrac{d}{dt}[\ln(4t)],\dfrac{d}{dt}[3\log_2(t)]\right) \\\\ &=\left(\dfrac{1}{t},\dfrac{3}{t\ln(2)}\right) \end{aligned}$ In conclusion, $f'(t)=\left(\dfrac{1}{t},\dfrac{3}{t\ln(2)}\right)$.